ODE and Population Models 2008 REU ODE and Population Models

Differential Equations! Intro Often know how populations change over time (e.g. birth rates, predation, etc.), as opposed to knowing a ‘population function’ Differential Equations! Knowing how population evolves over time w/ initial population  population function

Example – Hypothetical rabbit colony lives in a field, no predators. Let x(t) be population at time t; Want to write equation for dx/dt Q: What is the biggest factor that affects dx/dt? A: x(t) itself! more bunnies  more baby bunnies

1st Model—exponential, Malthusian Solution: x(t)=x(0)exp(at)

Critique Unbounded growth Non integer number of rabbits Unbounded growth even w/ 1 rabbit! Let’s fix the unbounded growth issue dx/dt = ????

Logistic Model dx/dt = ax(1-x/K) K-carrying capacity we can change variables (time) to get dx/dt = x(1-x/K) Can actually solve this DE Example: dx/dt = x(1-x/7)

Solutions: Critique: Still non-integer rabbits Still get rabbits with x(0)=.02

Fixed Points (equilibria) In Previous example: x=0 and x=7 are fixed points Fixed Point: dx/dt = 0 (so it’s fixed!) Stability: stable – near solutions tend to fixed point unstable = not stable

Stability Note: near x=7 d/dx ( du/dt) <0 (stable)

Stability Note: near x=0 d/dx ( du/dt) > 0 (unstable)

Taylor series at x* dx/dt=f(x) (no dependence on t) dx/dt = f(x)= c0+c1(x-x*)+c2(x-x*)^2+ …. (c0 = 0) If c1≠0, we can tell stability.

Moral: If dx/dt = f(x) and f(x*)=0 1) d/dx( f(x)) <0 at x* then x* is stable. 2) d/dx( f(x) ) >0 at x* then x* is unstable.

x’ versus x For first order autonomous equations, plotting x’ versus x encapsulates all this info x’ positive (unstable) x’ negative (stable)

Reality check Find and classify all equilibria of dx/dt = sin (x(t)) Firefly example (tomorrow)

Rabbit vs. Deer http://www.dcnr.state.pa.us/polycomm/pressrel/presqueislesp1100.htm

compete for the same food source. Let x(t) rabbits and y(t) deer compete for the same food source. dx/dt = dy/dt = Ax(1-x/K) -Cxy By(1-y/W) -Dxy Or…. (after changes of coordinates…) dx/dt = x(1-x-ay) dy/dt = y(b-by-cx)

Analysis of one case dx/dt = x(1-x-2y) dy/dt = y(2-2y-5x) Equlibria/Fixed Points: (0,0) , (0,1), (1,0), (1/4,3/8) Q: How do we know if these are stable or unstable? A: Linear approximation (derivative)

Linear Systems dx/dt= Ax (given by matrix mult) Fixed Point(s)?

What’s an eigenvalue again? Ax = λx (λ,x) are eigenvalue eigenvector pair Who cares? Think about: x(t) = exp (λt)x (Handout/Maple)

Other Tools Trapping regions Poincare Bendixson Nullclines Series solutions ,etc. Invariant Sets Bifurcations

Suppose we have 2 species; one predator y(t) (e. g Suppose we have 2 species; one predator y(t) (e.g. wolf) and one its prey x(t) (e.g. hare)

Actual Data

Model Want a DE to describe this situation dx/dt= ax-bxy = x(a-by) dy/dt=-cy+dxy = y(-c+dx) Let’s look at: dx/dt= x(1-y) dy/dt=y(-1+x)

Called Lotka-Volterra Equation, Lotka & Volterra independently studied this post WW I. Fixed points: (0,0), (c/d,a/b) (in example (1,1)).

Phase portrait y (1,1) x

A typical portrait: a ln y – b y + c lnx – dx=C

Solution vs time

Critiques Nicely captures periodic nature of data Orbits are all bounded, so we do not need a logistic term to bound x. Periodic cycles not seen in nature

Previte’s Population Projects 3-species chains - 2000 REU 3 Competing Species 2002/3 REU 4-species chains - 2004/5 REUs Adding a scavenger 2005/7 REUs (other interactions possible!)

3-species model (REU 2000) 3 species food chain! x = worms; y= robins; z= eagles dx/dt = ax-bxy =x(a-by) dy/dt= -cy+dxy-eyz =y(-c+dx-ez) dz/dt= -fz+gyz =z(-f+gy)

Critical analysis of 3-species chain ag > bf → unbounded orbits ag < bf → species z goes extinct ag = bf → periodicity

ag ≠ bf ag=bf

2000 REU and paper

Tools used in analysis Linearization Trapping regions Invariant sets Liapunov functions (“energy” functions)

One open conjecture ag>bf y tends to a limit as time increases all numerical evidence shows this, but no analytic proof.

4-species model dw/dt = aw-bxw =w(a-bx) dx/dt= -cx+dwx-exy =x(-c+dw-ey) dy/dt= -fy+gxy - hyz =y(-f+gx-hz) dz/dt= -iz+jyz =z(-i+jy)

2004/5 REU did analysis Orbits bounded again as in n=2 Quasi periodicity (next slide) ag<bf gives death to top 2 ag=bf gives death to top species ag>bf gives quasi-periodicity

Quasi-periodicity

Previte’s doughnut conjecture (ag>bf)

This is wide open Project never finished Proof seems too hard, may involve deep topics such as KAM theory, Hamiltonian systems

Simple Scavenger Model lynx beetle hare

Semi-Simple scavenger– Ben Nolting 2005 Know (x,y) -> (c, 1-bc) use this to see fc+gc+h=e every solution is periodic fc+gc+h<e implies z goes extinct fc+gc+h>e implies z to a periodic on the cylinder

Ben Nolting and his poster in San Antonio, TX

Scavenger Model with feedback (Malorie Winters & James Greene 2006/7)

Biological Example (crowding prey) crayfish Predator of mayfly nymph Scavenger of trout carcasses Rainbow Trout (predator) Mayfly nymph (Prey) Crayfish are scavenger & predator

Analysis (Malorie Winters) Regions of periodic behavior and Hopf bifurcations and stable coexistence. Regions with multi stability and dependence on initial conditions

Malorie Winters, and in New Orleans, LA

REU 2007 James Greene finds a model that exhibits chaos

2007 scavenger system dx/dt=x(1-bx-y-z) b, c, e, f, g, β > 0 dy/dt=y(-c+x) dz/dt=z(-e+fx+gy-βz)

Period Doubling cascade and attractor

TO DO Finish up the analysis from 2007 Including Hopf Bifurcation analysis, boundedness of orbits, and compare onset of chaos with other models Crowd the predator